systems analysis and controlcontrol.asu.edu/classes/mmae443/443lecture22.pdfassume we can plot the...

Systems Analysis and Control

Matthew M. PeetIllinois Institute of Technology

Lecture 22: The Nyquist Criterion

Overview

In this Lecture, you will learn:

Complex Analysis

• The Argument Principle

• The Contour Mapping Principle

The Nyquist Diagram

• The Nyquist Contour

• Mapping the Nyquist Contour

• The closed Loop

• Interpreting the Nyquist Diagram

M. Peet Lecture 22: Control Systems 2 / 30

Review

Recall: Frequency Response

Input:

u(t) =M sin(ωt+ φ)

Output: Magnitude and Phase Shift

y(t) = |G(ıω)|M sin(ωt+ φ+ ∠G(ıω))

0 2 4 6 8 10 12 14 16 18 20−2

−1.5

−1

−0.5

0

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1

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Linear Simulation Results

Time (sec)A

mpl

itude

Frequency Response to sinωt is given by G(ıω)

M. Peet Lecture 22: Control Systems 3 / 30

Review

Recall: Bode PlotThe Bode Plot is a way to visualize G(ıω):

1. Magnitude Plot: 20 log10 |G(ıω)| vs. log10 ω

2. Phase Plot: ∠G(ıω) vs. log10 ω

M. Peet Lecture 22: Control Systems 4 / 30

Bode Plots

If we only want a single plot we can use ω as a parameter.

−0.6 −0.4 −0.2 0 0.2 0.4 0.6

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

Nyquist Diagram

Real Axis

Imag

inar

y A

xis

A plot of Re(G(ıω)) vs. Im(G(ıω)) as a function of ω.

• Advantage: All Information in a single plot.

• AKA: Nyquist Plot

Question: How is this useful?

M. Peet Lecture 22: Control Systems 5 / 30

The Nyquist Plot

To Understand Nyquist:

• Go back to Root Locus

• Consider a single zero: G(s) = s.

Draw a curve around the poleWhat is the phase at a point on thecurve?

∠G(s) = ∠s

< s

M. Peet Lecture 22: Control Systems 6 / 30

The Nyquist Plot

Consider the phase at four points

1. ∠G(a) = ∠1 = 0◦

2. ∠G(b) = ∠− ı = −90◦

3. ∠G(c) = ∠− 1 = −180◦

4. ∠G(d) = ∠ı = −270◦

The phase decreases along the curveuntil we arrive back at a.

• The phase resets at a by +360◦

a

b

c

d

The reset is Important!

• There would be a reset for any closed curve containing z or any startingpoint.

• We went around the curve Clockwise (CW).I If we had gone Counter-Clockwise (CCW), the reset would have been

−360◦.

M. Peet Lecture 22: Control Systems 7 / 30

The Nyquist Plot

Now consider the Same Curve with

G(s) = s+ 2

Phase at the same four points

1. ∠G(a) = ∠3 = 0◦

2. ∠G(b) = ∠2− ı ∼= −30◦

3. ∠G(c) = ∠1 = 0◦

4. ∠G(d) = ∠2 + ı ∼= 30◦a

b

c

d

30o

In this case the transition back to 0◦ is smooth.

• No reset is required!

M. Peet Lecture 22: Control Systems 8 / 30

The Nyquist Plot

Question What if we had encircled 2 zeros?

Phase at the same four points

1. ∠G(a) = 0◦

2. ∠G(b) = −180◦

3. ∠G(c) = −360◦

4. ∠G(d) = −540◦

• The phase resets at a by +720◦

a

b

c

d

60o120o

Rule: The reset is +360 ·#zeros.

M. Peet Lecture 22: Control Systems 9 / 30

The Nyquist Plot

Question What about encircling a pole?

Consider the phase at four points

1. ∠G(a) = ∠1 = 0◦

2. ∠G(b) = ∠ 1−ı = ∠ı = 90◦

3. ∠G(c) = ∠− 1 = 180◦

4. ∠G(d) = ∠ 1ı = ∠− ı = 270◦

• The phase resets at a by −360◦

a

b

c

d

<s-1 = 90o

Rule: The reset is −360 ·#poles.

M. Peet Lecture 22: Control Systems 10 / 30

The Nyquist Plot

Question: What if we combine a pole and a zero?

Consider the phase at four points

1. ∠G(a) = 0◦

2. ∠G(b) = −60◦

3. ∠G(c) = 0◦

4. ∠G(d) = 60◦

• There is no reset at a.

a

b

c

d

120o

<s-1 = -60o

Rule: Going CW, the reset is +360 · (#zeros −#poles).

A consequence of the Argument Principle from Complex Analysis.

M. Peet Lecture 22: Control Systems 11 / 30

The Nyquist Plot

How can this observation be used?Consider Stability.

• G(s) is stable if it has no poles in the right half-plane

Question: How to tell if any poles are in the RHP?Solution: Draw a curve around the RHP and count the resets.

Define the Curve:

• Starts at the origin.

• Travels along imaginary axis till r =∞.

• At r =∞, loops around clockwise.

• Returns to the origin along imaginary axis.

A Clockwise Curve

The reset is +360 · (#zeros −#poles).

If there is a negative reset, there is a pole in theRHP

r = ∞

M. Peet Lecture 22: Control Systems 12 / 30

The Nyquist Plot

If we encircle the right half-plane,

The reset is +360 · (#zeros −#poles).

Question 1:

• How to determine the number ofresets along this curve?

Question 2:

• Zeros can hide the poles!

• What to do?

r = ∞

resets = 0

M. Peet Lecture 22: Control Systems 13 / 30

Contour Mapping

Lets answer the more basic question first:

• How to determine the number of resets along this curve?

Definition 1.

Given a contour, C ⊂ X, and a function G : X → X, the contour mappingG(C) is the curve {G(s) : s ∈ C}.

In the complex plane, we plot

Im(G(s)) vs. Re(G(s))

along the curve C• Yields a new curve, CG.

s

G(s)

Im( G(s) )

Re( G(s) )

M. Peet Lecture 22: Control Systems 14 / 30

Contour Mapping

Key Point: For a point on the mapped contour, s∗ = G(s),

∠s∗ = ∠G(s)

• We measure θ, not phase.

To measure the 360◦ resets in ∠G(s)

• We count the number of +360◦ resets in θ!

• We count the number of times CG encircles the origin Clockwise.

A reset occurs every time CG encirclesthe origin clockwise.

• Makes the resets much easier tocount!

Assumes the contour doesn’t hit anypoles or zeros, otherwise

• G(s)→∞ and we lose count.

• G(s)→ 0 and we lose count.

s

s*= G(s)

θ = < G(s)

M. Peet Lecture 22: Control Systems 15 / 30

Contour MappingExample 1

Remember Direction is importantIf the original Contour was counter-clockwise

The reset is +360 · (#poles −#zeros).

In this example we see CG• encircles the origin once going

ClockwiseI θa = 0I θb = −90◦

I θc = −180◦

I θd = −270◦

• A Positive Reset of +360◦.

a

b

c

d

Thus +360 · (#poles −#zeros) = +360

(#poles −#zeros) = 1

At least one pole in the region.M. Peet Lecture 22: Control Systems 16 / 30

Contour Mapping

Assume the original Contour was clockwise

The reset is +360 · (#zeros −#poles).

There are 5 counter-clockwiseencirclements of the origin.

• A Negative Reset of −360◦ · 5.

Thus

+360 · (#zeros −#poles) = −360 · 5

(#zeros −#poles) = −5

At least 5 poles in the region.

M. Peet Lecture 22: Control Systems 17 / 30

The Nyquist Contour

Conclusion: If we can plot the contour mapping, we can find the relative # ofpoles and zeros.

Definition 2.

The Nyquist Contour, CN is a contour which contains the imaginary axis andencloses the right half-place. The Nyquist contour is clockwise.

A Clockwise Curve

• Starts at the origin.

• Travels along imaginary axis till r =∞.

• At r =∞, loops around clockwise.

• Returns to the origin along imaginary axis.

r = ∞

M. Peet Lecture 22: Control Systems 18 / 30

The Nyquist Contour

To map the Nyquist Contour, we deal with twoparts

• The imaginary Axis.

• The loop at ∞.

The Imaginary Axis

• Contour Map is G(ıω)

• Plot Re(G(ıω)) vs. Im(G(ıω))

Data Comes from Bode plot

• Plot Re(G(ıω)) vs. Im(G(ıω))

Map each point on Bode to a point on Nyquist

s*= G(iω)

θ = < G(iω){ r = |G(iω)|

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0

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Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

M. Peet Lecture 22: Control Systems 19 / 30

The Nyquist Contour

The Loop at ∞: 2 Cases

G(s) =n(s)

d(s)=a0s

m + · · · amb0sn + · · · bn

Case 1: G(s) is Proper, but notstrictly

• Degree of d(s) same as n(s)

• As ω →∞, G(s) becomesconstant

I Magnitude becomes fixed

lims→∞

n(s)

d(s)=n(s)

d(s)=a0b0

I Phase varies (More on this later)

We can use the Nyquist Plot

−10

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Mag

nitu

de (

dB)

100

101

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104

105

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

M. Peet Lecture 22: Control Systems 20 / 30

The Nyquist Contour

The Loop at ∞:

G(s) =n(s)

d(s)=a0s

m + · · · amb0sn + · · · bn

Case 1: G(s) is Strictly Proper

• Degree of d(s) greater than n(s)

• As ω →∞, |G(ıω)| → 0

lims→∞

G(s) = limω→∞

n(s)

d(s)= 0

−40

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0

5

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Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

Can’t tell what goes on at ∞!

Nyquist Plot is useless

M. Peet Lecture 22: Control Systems 21 / 30

The Nyquist Contour

If we limit ourself to proper, but not strictly proper.

Because the Nyquist Contour is clockwise,

The number of clockwise encirclements of 0 is

• The #poles −#zeros in the RHP

Conclusion: Although we can map the RHP onto the Nyquist Plot, we havetwo problems.

• Can only determine #poles −#zeros

• Doesn’t work for strictly proper systems.

M. Peet Lecture 22: Control Systems 22 / 30

The Nyquist Contour

Our solution to all problems is to consider Systems in Feedback

• Assume we can plot the Nyquist plot for the open loop.

• What happens when we close the loop?

The closed loop iskG(s)

1 + kG(s)

We want to know when1 + kG(s) = 0

Question: Does 1k +G(s) have any zeros in the RHP?

G(s)k+

-

y(s)u(s)

M. Peet Lecture 22: Control Systems 23 / 30

The Nyquist ContourClosed Loop

This is a better question.1k +G(s) is Proper, but not Strictly

1

k+G(s) =

d(s) + kn(s)

kd(s)

• Degree of d(s) greater than or equals n(s)

• degree(d(s) + kn(s)) = degree(d(s))

Numerator and denominator have same degree!

We know about the poles of 1k +G(s)

• poles are the poles of the open loop

• We know if the open loop is stable!

• we know if any poles are in RHP.

M. Peet Lecture 22: Control Systems 24 / 30

The Nyquist ContourClosed Loop

Mapping the Nyquist contour of 1k +G(s) is easy!

1. Map the Contour for G(s)

2. Add 1k to every point

1/k

Shifts the plot by factor 1k

M. Peet Lecture 22: Control Systems 25 / 30

The Nyquist ContourClosed Loop

Conclusion: If we map the Nyquist Contour for 1k +G(s)

• The # of clockwise encirclements of 0 is #poles −#zeros of closed loop inthe RHP.

• The # of zeros of 1k +G(s) in RHP is # of clockwise encirclements plus #

of open-loop poles of G(s) in RHP.

Instead of shifting the plot, we can shift theorigin to point − 1

k

The number of unstable closed-loop poles isN + P , where

• N is the number of clockwiseencirclements of −1k .

• P is the number of unstable open-looppoles.

If we get our data from Bode, typically P = 0-1/k

M. Peet Lecture 22: Control Systems 26 / 30

The Nyquist ContourExample

-1/k

Two CCW encirclements of − 1k

• Assume 1 unstable Open Loop pole P = 1• Encirclements are CCW: N = −2• N + P = −1: No unstable Closed-Loop Poles

M. Peet Lecture 22: Control Systems 27 / 30

The Nyquist ContourExample

Nyquist lets us quickly determine the regions of stability

The Suspension Problem

• Open Loop is Stable: P = 0

• No encirclement of −1/kI Holds for any k > 0

Closed Loop is stable for any k > 0.

−4 −2 0 2 4 6−8

−6

−4

−2

0

2

4

6

8

Nyquist Diagram

Real Axis

Imag

inar

y A

xis

M. Peet Lecture 22: Control Systems 28 / 30

The Nyquist ContourExample

The Inverted Pendulum with Derivative Feedback

• Open Loop is Unstable: P = 1

• CCW encirclement of −1/kI Holds for any −2 < −1

k< 0

I Holds for any k > 12

• When k ≥ 12 , N = −1

Closed Loop is stable for k > 12 .

−2.5 −2 −1.5 −1 −0.5 0 0.5−1

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−0.2

0

0.2

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1

Nyquist Diagram

Real Axis

Imag

inar

y A

xis

M. Peet Lecture 22: Control Systems 29 / 30

Summary

What have we learned today?

Complex Analysis

• The Argument Principle

• The Contour Mapping Principle

The Nyquist Diagram

• The Nyquist Contour

• Mapping the Nyquist Contour

• The closed Loop

• Interpreting the Nyquist Diagram

Next Lecture: Drawing the Nyquist Plot

M. Peet Lecture 22: Control Systems 30 / 30